Friction and Rolling Analytical Question

[modulus]

I found this question in HRW (sixth edition), one of the 'checkpoints' (checkpoint 2, to be precise) in chapter 12 on 'rolling, torque, and angular momentum':

Homework Statement

Disks A and Bare identical and roll across a floor with equal speeds. The disk A roll up an incline, reaching a maximum height h, and the disk B moves up an incline that is identical except that it is frictionless. Is the maximum height reached by disk B greater than, less than, or equal to h?

Homework Equations

Work - Kinetic Energy Theorem
Concept of External Forces and Energy Changes (Potential and Kinetic) in a System
Newton's Second Law for Rotation

The Attempt at a Solution

I assumed the disks to lose all their kinetic energy at the maximum height (no more rotation, and no more translation).
So, for disk A, the (negative) work done by gravity (which comes out to be equal to mgh), and the (positive) work done by friction will equal the change in kinetic energy (which should be equal to 0.5{I[w]^2 + m[v]^2}, the rotational and translational kinetic energy.
For disk B, the equation will be [mgh = rotational + translational kinetic energy].
Because of the term pertaining to friction in A's equation, A's height comes out to be greater.

But, in the sample problem immediately after that, it explains that frictional force cannot do any work on a body rolling smoothly down an incline (what?!). But if this is true, friction would never make any difference to a body's rolling motion, so both should reach the same height...yet, the correct answer is that A will reach a greater height.

How?

Please help.

 

[PeterO]

I found this question in HRW (sixth edition), one of the 'checkpoints' (checkpoint 2, to be precise) in chapter 12 on 'rolling, torque, and angular momentum':

Homework Statement

Disks A and Bare identical and roll across a floor with equal speeds. The disk A roll up an incline, reaching a maximum height h, and the disk B moves up an incline that is identical except that it is frictionless. Is the maximum height reached by disk B greater than, less than, or equal to h?

Homework Equations

Work - Kinetic Energy Theorem
Concept of External Forces and Energy Changes (Potential and Kinetic) in a System
Newton's Second Law for Rotation

The Attempt at a Solution

I assumed the disks to lose all their kinetic energy at the maximum height (no more rotation, and no more translation).
So, for disk A, the (negative) work done by gravity (which comes out to be equal to mgh), and the (positive) work done by friction will equal the change in kinetic energy (which should be equal to 0.5{I[w]^2 + m[v]^2}, the rotational and translational kinetic energy.
For disk B, the equation will be [mgh = rotational + translational kinetic energy].
Because of the term pertaining to friction in A's equation, A's height comes out to be greater.

But, in the sample problem immediately after that, it explains that frictional force cannot do any work on a body rolling smoothly down an incline (what?!). But if this is true, friction would never make any difference to a body's rolling motion, so both should reach the same height...yet, the correct answer is that A will reach a greater height.

How?

Please help.

When you say

"So, for disk A, the (negative) work done by gravity (which comes out to be equal to mgh), and the (positive) work done by friction will equal the change in kinetic energy"

I am not sure why you said one work was positive, and one was negative? Surely the disc stops because it does work against gravity [by going up] and friction will ALWAYS be work against what ever you are try to do. Friction will NEVER do work to help you along - friction opposes all changes.

 

[ehild]

Rolling happens when static friction is great enough to keep the point in contact with ground in momentary rest. The translational speed of the rolling body is equal to the linear speed of the rim. If the body stops, both rotation and translation are stopped. But static friction does not do any work, as there is no displacement for the force.
Without friction, there is no rolling. The disk reaching the slope rotating, will rotate with the original angular velocity, as there is no torque to change it.

 

[ehild]

Surely the disc stops because it does work against gravity [by going up] and friction will ALWAYS be work against what ever you are try to do. Friction will NEVER do work to help you along - friction opposes all changes.

It is a bit strange to say that the disk does work against gravity. On what does it do work? There is only one thing it is contact with: the slope and as it is fixed to the Earth, the disk does work on Earth.

As for friction to be help or not, think of a heavy load on the plate of a lorry, and the vehicle accelerating. What would happen to the load without friction?

ehild

 

[modulus]

Thank you so much, ehild and PeterO. Your help, along with this amazingly helpful module I found on the net: http://cnx.org/content/m14385/latest/
have clarified my doubts.

Ehild, you wrote:
"Without friction, there is no rolling. The disk reaching the slope rotating, will rotate with the original angular velocity, as there is no torque to change it."
I believe this is with reference to disk B, right? If so, then I've got everything right...thank you so much, again!

 

[ehild]

Ehild, you wrote:
"Without friction, there is no rolling. The disk reaching the slope rotating, will rotate with the original angular velocity, as there is no torque to change it."
I believe this is with reference to disk B, right? If so, then I've got everything right...thank you so much, again!

Yes, it refers to disk B. The whole KE of A will transform into PE, but only the translational KE transforms into PE in case of disk B.
When disk A rolls uphill, both its translational and rotational motion decelerates. The force of static friction (Fs) provides a torque that opposes rotation, that means it points upward along the slope, so it is an accelerating force for the translational motion opposite to the component of gravity. Therefore the deceleration is slower than without friction, the disk moves longer and travels a longer distance than disk B.



ehild